A053831 Sum of digits of n written in base 11.
0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16, 7, 8, 9, 10, 11, 12, 13, 14, 15
Offset: 0
Examples
a(20) = 1 + 9 = 10 because 20 is written as 19 base 11.
Links
- Tanar Ulric, Table of n, a(n) for n = 0..10000
- Jeffrey O. Shallit, Problem 6450, Advanced Problems, The American Mathematical Monthly, Vol. 91, No. 1 (1984), pp. 59-60; Two series, solution to Problem 6450, ibid., Vol. 92, No. 7 (1985), pp. 513-514.
- Robert Walker, Self Similar Sloth Canon Number Sequences.
- Eric Weisstein's World of Mathematics, Digit Sum.
Crossrefs
Programs
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C
int Base11DigitSum(int n) { int count = 0; while (n != 0) { count += n % 11; n = n / 11; } return count; } // Tanar Ulric, Oct 20 2021
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Mathematica
Table[Plus @@ IntegerDigits[n, 11], {n, 0, 86}] (* or *) Nest[ Flatten[ #1 /. a_Integer -> Table[a + i, {i, 0, 10}]] &, {0}, 2] (* Robert G. Wilson v, Jul 27 2006 *)
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PARI
a(n)=if(n<1,0,if(n%11,a(n-1)+1,a(n/11)))
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PARI
a(n)=sumdigits(n,11) \\ Charles R Greathouse IV, Oct 20 2021
Formula
From Benoit Cloitre, Dec 19 2002: (Start)
a(0)=0, a(11n+i) = a(n)+i for 0 <= i <= 10.
a(n) = n-(m-1)*(Sum_{k>0} floor(n/m^k)) = n-(m-1)*A064458(n). (End)
a(n) = A138530(n,11) for n > 10. - Reinhard Zumkeller, Mar 26 2008
Sum_{n>=1} a(n)/(n*(n+1)) = 11*log(11)/10 (Shallit, 1984). - Amiram Eldar, Jun 03 2021
Comments